For each rational $q$ there is an integer $n$ such that $q+n=271$, and vice versa

Question

For all $q \in \Bbb Q$ there exists $n \in \Bbb Z$ so that $q + n = 271$.

This is true? Because both $q$ and $n$ are rational numbers and $271$ is an integer thus it's a rational number?

Also,

For all $n \in \Bbb Z$ there exists $q \in \Bbb Q$ so that $q + n = 271$.

Isn't it the same thing? One should be true and one should be false. Any help?

Answer 1

For all $q \in \Bbb Q$ there exists $n \in \Bbb Z$ such that $q + n = 271$.

False.

$\Bbb Q$ is the set of rationals, i.e. fractions: $\Bbb Q = \{\ldots, -\frac 1 3,- \frac 2 2,-\frac 1 2, 0, \frac 1 2, \frac 2 2, \frac 1 3, \ldots\}$.

$\Bbb Z$ is the set of integers: $\{\ldots, -3, -2, -1, 0, 1,2,3,\ldots\}$.

Take $q = \frac 1 2$ for an easy counterexample. Generally, any $q \in \Bbb Q$ such that $q \notin \Bbb Z$ yields a counterexample to the claim.

For all $n \in \Bbb Z$ there exists $q \in \Bbb Q$ such that $q + n = 271$.

True. It's easy to see by example why this is true, but there's a more general reason.

From the axioms of arithmetic, we know that for any $n \in \Bbb Z$ there exists an $m \in \Bbb Z$ such that $n+m = 271$. We also know that $\Bbb Z \subset \Bbb Q$, i.e. that the integers are a subset of the rationals. So when we go to pick the second integer $m$, that integer $m$ is also necessarily contained in the rationals (as $\frac m 1$).

Answer 2

No, it's not the same thing.

Note that $\frac{1}{2}$, a rational number, cannot be added to any integer to get 271. The first one is false.

The second is true. For any integer, there is some integer I can add to it to yield 271, and all integers are also rational numbers.